//Given a linked list, swap every two adjacent nodes and return its head. You mu
//st solve the problem without modifying the values in the list's nodes (i.e., onl
//y nodes themselves may be changed.) 
//
// 
// Example 1: 
//
// 
//Input: head = [1,2,3,4]
//Output: [2,1,4,3]
// 
//
// Example 2: 
//
// 
//Input: head = []
//Output: []
// 
//
// Example 3: 
//
// 
//Input: head = [1]
//Output: [1]
// 
//
// 
// Constraints: 
//
// 
// The number of nodes in the list is in the range [0, 100]. 
// 0 <= Node.val <= 100 
// 
// Related Topics 递归 链表 
// 👍 914 👎 0


package leetcode.editor.cn;

//Java：Swap Nodes in Pairs
 class P24SwapNodesInPairs {
    public static void main(String[] args) {
        Solution solution = new P24SwapNodesInPairs().new Solution();
        // TO TEST
    }
    //leetcode submit region begin(Prohibit modification and deletion)

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     * int val;
     * ListNode next;
     * ListNode() {}
     * ListNode(int val) { this.val = val; }
     * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
     * }
     */
    class Solution {
        public ListNode swapPairs(ListNode head) {
            if (head == null) {
                return null;
            }
            ListNode a, b;
            a = b = head;
            for (int i = 0; i < 2; i++) {
                if (b == null) {
                    return head;
                }
                b = b.next;
            }
            ListNode newHead = reverse(a, b);
            a.next = swapPairs(b);

            return newHead;
        }

        ListNode reverse(ListNode a, ListNode b) {
            ListNode pre = null, cur = a, nxt;
            while (cur != b) {
                nxt = cur.next;
                cur.next=pre;
                pre = cur;
                cur = nxt;
            }
            return pre;
        }

    }
//leetcode submit region end(Prohibit modification and deletion)

}